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Georgia [21]
2 years ago
10

……………………………………………………………..?

Physics
1 answer:
TEA [102]2 years ago
4 0

Answer:

last is the answer.

increase the voltage in order to send energy faster.

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A snail moves 40 cm in<br> 20 seconds. How fast<br> did it move?
nasty-shy [4]

Answer:

too fast to be a snail

......

2 cm per second

5 0
3 years ago
This is an electrical device that changes the voltage of electricity through the use of two set of coils
Fynjy0 [20]
Voltmeter is the right answer
3 0
3 years ago
Read 2 more answers
A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

8 0
3 years ago
Calculate the radius of each ball to the nearest tenth. Use the formula r = d/2 where r is the radius and d is the diameter. Rec
erma4kov [3.2K]

Answer:

What is the radius of the table tennis ball?

 ⇒ 2.1 cm  

What is the radius of the golf ball?

2.0  cm

Explanation:

divide the radius and round it to the nearest 10th place..... but hope that help ;)

8 0
3 years ago
Read 2 more answers
I dont know how to do this at all please help
worty [1.4K]
Wow !  I understand your shock.  I shook and vibrated a little
when I looked at this one too.

The reason for our shock is all the extra junk in the question,
put there just to shock and distract us.

"Neutron star", "5.5 solar masses", "condensed burned-out star".
That's all very picturesque, and it excites cosmic fantasies in
out brains when we read it, but it's just malicious decoration.
It only gets in the way, and doesn't help a bit.

The real question is:

What is the acceleration of gravity 2000 m from
the center of a mass of 1.1 x 10³¹ kg ?

Acceleration of gravity is

                           G  ·  M / R²

      =  (6.67 x 10⁻¹¹ N·m²/kg²) · (1.1 x 10³¹ kg) / (2000 m)²

      =  (6.67 x 10⁻¹¹  ·  1.1 x 10³¹ / 4 x 10⁶)      (N) · m² · kg / kg² · m²

      =             1.83 x 10¹⁴           (kg · m / s²) · m² · kg / kg² · m²

      =             1.83 x 10¹⁴            m / s²      

That's about  1.87 x 10¹³  times the acceleration of gravity on
Earth's surface.

In other words, if I  were standing on the surface of that neutron star,
I would weigh  1.82 x 10¹² tons, give or take.     
3 0
3 years ago
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